# Infinite product of non-units is always 0

Is there an elementary way to see that a product of an infinite number of non-units in a (noncommutative) ring is always $$0$$? Here, by non-units I mean elements that are not left-invertible.

This statement should be equivalent to every principal left ideal being Noetherian. Since the latter is true for any ring, the former must also be true, but I couldn't find an elementary proof.

• Infinite products only make sense in topological rings. This is because a binary pruducf can only be extended to finite products, algebraically. Oct 30, 2022 at 20:36
• It is not true in $\mathbb Z$ with the usual topology. But it is true if you add the condition "If the infinite product of non-units converges, it converges to $0.$" Not sure if that would be true in general, though. Oct 30, 2022 at 20:41
• It seems unlikely to be true. For example, in the ring $R=\mathbb Z[1/2]$ with the usual topology, I think we can fairly easily find non-units whose product is $1.$ Oct 31, 2022 at 0:04

For example, the ring: $$\mathbb Z[1/2]=\left\{\frac a{2^k}\mid a,k\in\mathbb Z, k\geq 0\right\}$$ with the topology inherited from the rational numbers.
The units are $$\pm 2^j$$ for any $$j\in\mathbb Z.$$
Then the product: $$\frac{3}{4}\cdot \frac{5}{4}\cdot \frac{17}{16}\cdots \frac{2^{2^k}+1}{2^{2^k}}\cdots$$
converges to $$1,$$ but all of the terms are non-units. (The first term, $$\frac34,$$ does not match the general pattern, but the other terms do.)
A partial product, up to the term for $$k,$$ can be shown to be $$\frac{2^{2^{k+1}}-1}{2^{2^{k+1}}}$$ by induction.